PRIMARY RADICALS IN LATTICE MODULES

C S MANJAREKAR AND U N KANDALE

Abstract. In this paper, we define primary radical of an elementin a lattice module and obtain some properties of it and characterizep-primary elements of lattice modules.

Keywords: Multiplicative lattice, lattice module, prime element, pri-mary element.

1. Introduction

Multiplicative lattice is a complete lattice provided with commutative,associative and join distributive multiplication in which the largest elementI acts as a multiplicative identity. An element p ∈ L is called proper if p < 1.A proper element a ∈ L is called prime if ab 6 p ⇒ a 6 p or b 6 p for a, b ∈L. A proper element p ∈ L is said to be primary if ab 6 p ⇒ a 6 p or bn 6 p

for some n ∈ Z+. If a, b ∈ L, (a : b) is the join of all elements c in L suchthat cb 6 a. If a ∈ L,

√a = ∨{x ∈ L | xn 6 a, forsome n ∈ Z+}. Let

M be a complete lattice and L be a multiplicative lattice. Then M is calledL-module or module over L if there is a multiplication between elements ofL and elements of M written as aB where a ∈ L and B ∈M which satisfiesthe following conditions,

(1) (ab)B = a(bB)(2) (∨

αaα)(∨

βBβ) = ∨

αβaαBβ

(3) IB = B

(4) OB = OM for all a, b, aα ∈ L and B, Bβ ∈ B where 0 is the infimumof L and 0M is the infimum of M.Elements of L will generally bedenoted by a,b,c ... except that the least element of L will bedenoted by 0 and the greatest element of L will be denoted by I.Theelements of M will be denoted by A,B,C ... except that the leastelement and the greatest element of M wll be denoted by OM andIM respectively.If N ∈ M and a ∈ L then (N : a) = ∨{X ∈ M |aX 6 N}.If A, B ∈ M then (A : B) = ∨{x ∈ L | xB 6 A}. ForB ∈M,

√B = ∨{a ∈ L | anIM 6 B for some n ∈ Z+}.Let N be

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a proper element of a L-module M . N is called prime element of Mif for any a ∈ L and A ∈ M, aA 6 N implies A 6 N or aIM 6 N

i.e. a 6 (N : IM ). Let N be a proper element of a L-module M. N iscalled a primary element of M if for any a ∈ L and A ∈M, aA 6 N

implies A 6 N or anIM 6 N i.e an 6 (N : IM ). If N is a primeelement of M then (N : IM ) = p is a prime element of L and wesay that N is p-prime. Also if N is a primary element of M then(N : IM ) is primary element of L.Then, p =

√(N : IM ) is a prime

element and we say that N is p-primary.An element B 6= IM of Mis called a semiprime element if ∀a, b ∈ L such that abIM 6 B theneither aIM 6 B or bIM 6 B.For details see [1].

2. PRIME,PRIMARY AND SEMIPRIME ELEMENTS

Now we charactarize p-primary element in lattice modules.

Theorem 2.1. Let N be a proper element of a L-module M. Let p be aprime element of L.Then N is p-primary element of M if and only if

(1) p 6√

(N : IM )(2) cA N for all c ∈ L, c p, A ∈M, A N .

Proof. If N is p-primary then by definition p =√

(N : Im). Now, let c ∈L, c p and A ∈ M, A N . Let cA 6 N .Then ∃ a positive integer nsuch that cn 6 (N : IM ) i.e. c 6

√(N : IM ) = p (becauseA N) which

is a contradiction. Hence, cA N . Conversely, assume that, (1) and (2)hold. Let a ∈ L, A ∈ M, aA 6 N . Assume further that, A N .Then by(1) and (2), a 6 p 6

√(N : IM ), that is an 6 (N : IM ) forsome n ∈

Z+, anIM 6 N forsome n ∈ Z+. Therefore, N is primary element ofM. Next, we must show that, p =

√(N : IM ). Let, a 6

√(N : IM ) then

an 6 (N : IM ) for some positive integer n and so anIM 6 N . By hypothesis,anIM 6 N implies either an 6 p or IM 6 N . Since, N is proper, IM N

and an 6 p. But, p is prime element, hence a 6 p.So√

(N : IM ) 6 p. Butp 6

√(N : IM ) implies

√(N : IM ) = p. Therefore, N is p-primary. �

The next result gives the relation between semiprime element of a latticemodule M and multiplicative lattice L.The study of semiprime submodulesof a module is carried out by Tavallaee H.A.[3].

Theorem 2.2. Let M be an L-module. If N is a semiprime element of Mthen (N : IM ) is semiprime element of L.

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Proof. Suppose, N is semiprime element of M. We show that,√

(N : IM ) =(N : IM ).By befinition,

√(N : IM ) = ∨{x | xn 6 (N : IM )}.So, (N :

IM ) 6√

(N : IM ). Next, a 6√

(N : IM ) implies an 6 (N : IM ) forsome n ∈Z+ i.e. anIM 6 N . Since, N is semiprime, aan−1IM 6 N , either aIM 6 N

or an−1IM 6 N that is a 6 (N : IM ) or an−1 6 (N : IM ). If a 6 (N : IM )then

√(N : IM ) 6 (N : IM ). If an−1 6 (N : IM ) then an−1IM 6 N that is

aan−2IM 6 N implies aIM 6 N or an−2IM 6 N . Continuing in this way,we get, a 6 (N : IM ). Therefore,

√(N : IM ) 6 (N : IM ), and we have

(N : IM ) =√

(N : IM ) �

Now,we have the characterization of semiprime element of a lattice mod-ule M.

Theorem 2.3. Let M be a multiplication lattice module. An element N ofM is semiprime if and only if (N : IM ) is a semiprime element of L.

Proof. By lemma (2.1) in [1], (N : IM ) ∼= [(N : IM ), IL]. Thus, N is asemiprime element of M if and only if (N : IM ) is a semiprime element ofL. �

The following theorem gives the characterization of prime and primaryelements of a lattice module M over L in terms of prime and primary ele-ments of L.

Theorem 2.4. Let M be a multiplication L-module then(1) An element N of M is prime if and only if (N : IM ) is a prime

element of L.(2) An element N of M is primary if and only if (N : IM ) is a primary

element of L.

Proof. (1) Suppose, N is prime. To show that, (N : IM ) is prime element ofL let, ab 6 (N : IM ), where a, b ∈ L with b (N : IM ). We have, abIM 6

N . Since, N is prime and bIM N ,therefore, aIM 6 N i.e a 6 (N : IM ).Conversely, suppose (N : IM ) is a prime element of L.Let, abIM 6 N .Thenab 6 (N : IM ). Since, (N : IM ) is prime, a 6 (N : IM ) or b 6 (N : IM ),that is aIM 6 N or bIM 6 N . Hence, N is semiprime. Therefore, N isprime (by proposition (3.2), [1])

(2) Suppose, N is primary. We show that, (N : IM ) is a primary elementof L. Let, ab 6 (N : IM ). Then, abIM 6 N . Since N is primary, eitheraIM 6 N or bnIM 6 N for some positive integer n. That is, a 6 (N : IM )or bn 6 (N : IM ). Thus, (N : IM ) is a primary element of L. Conversely,let (N : IM ) be primary. To show that, N is primary. Let aB 6 N

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and let B N .Suppose, anIM N ∀n. Then, an (N : IM )∀n. So,an (N : IM )impliesan (N : B) that is a (N : B).Hence, aB N acontradiction. Thus, anIM 6 N . Therefore, N is primary. �

The next result gives the characterization of semiprime element in latticemodule M.

Theorem 2.5. Let, M be a L-module where M satisfies ACC and is amultiplication lattice module and let N be a proper element of M. Then N issemiprime if and only if it is a radical element.

Proof. Suppose, N is semiprime. We show that,√

N =√

(N : IM ) = (N :IM ).We have, (N : IM ) 6

√(N : IM ). Let, x ∈ L be such that xnIM 6

N thatis(xIM )(xn−1IM ) 6 N . If n = 2, the result is obvious. Let, n >

2, xIM 6 N or xn−1IM 6 N . If xn−1IM 6 N then xIM 6 N or xn−2IM 6

N. Continuing in this way, xIM 6 N which implies x 6 (N : IM ). Thus,√(N : IM ) 6 (N : IM ) and N is a radical element. Conversely, assume

that N is a radical element. Since N is a radical element then (N : IM )is semiprime element of L. By the theorem 2.3, N is semiprime element ofM. �

3. Primary radicals

The concept of the primary radical of a submodule is introduced by LamisJ.M.A.[2].We generalize this concept for lattice modules and obtain someproperties of it. Let M be a L-Module and N be any element of L. We defineprimary radical of L-Module M to be the meet of all primary elements of M.The prime radical of an element N of M is defined as the meet of all primeelements of M containing N. If there is no prime element of M containingN, we write radM (N) = IM . The primary radical of an element N of Mdenoted by pradM (N) is defined as the meet of all primary elements of Mcontaining N. If there is no primary element of M containing N then wewrite pradM (N) = IM .

Theorem 3.1. Let, N1, N2 and N be elements of L-Module M then,(1) N 6 pradM (N)(2) pradM (N1 ∧N2) 6 pradM (N1) ∧ pradM (N2)(3) pradM (pradM (N)) = pradM (N).

Proof. (1) Obvious.(2) pradM (N1 ∧ N2) =∧{Pi | Pi is primary and N1 ∧ N2 6 Pi}6 [∧{Pi is primary | N1 6 Pi}] ∧ [∧{Pi is primary | N2 6 Pi}]= pradM (N1) ∧ pradM (N2).

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(3) prad(pradN) = prad[∧{Pi is primary | N 6 Pi}] = ∧{Pi |[∧{Pi is primary | N 6 Pi}] 6 Pi} = ∧{Pi is primary | N 6 Pi}= pradN .

�

An element K of M is said to be meet irreducible if N1 ∧N2 6 K implieseither N1 6 K or N2 6 K where N1 and N2 are elements of M.

Corollary 3.1. If every primary element of M is meet irreducible thenpradM (N1 ∧N2) = pradM (N1) ∧ pradM (N2).

Theorem 3.2. Let N1 and N2 be elements of L-module of M such that√(N1 : IM ) ∨

√(N2 : IM ) = 1 then pradM (N1 ∧ N2) = pradM (N1) ∧

pradM (N2).

Proof. If K is a primary element containing N1∧N2 then√

(N1 ∧N2 : IM ) 6√(K : IM ).We have,.

√(N1 : IM ) ∧

√(N2 : IM ) = [∨{x ∈ L | xn 6 (N1 :

IM )}] ∧ [∨{x ∈ L | xn 6 (N2 : IM )}].Also,√

(N1 ∧N2 : IM ) = ∨{x ∈ L |xn 6 (N1 ∧ N2) : IM}. Let x ∈ L.Then, xn 6 (N1 : IM ), xm 6 (N2 : IM ),for some positive integers m,n. If n > m, xn 6 (N1 : IM ), xn 6 (N2 : IM ).Therefore,

√(N1 : IM )∧

√(N2 : IM ) 6

√(N1 ∧N2 : IM ). If K is primary,

(K : IM ) is a primary element of L and hence√

(K : IM ) is prime elementof L.Therefore,

√(N1 : IM ) 6

√(K : IM ) or

√(N2 : IM ) 6

√(K : IM ).If√

(N1 : IM ) 6√

(K : IM ) then√

(N2 : IM ) √

(K : IM ). Because, 1 =√(N1 : IM ) ∨

√(N2 : IM ) 6

√(K : IM ) implies IM 6 K and hence K =

IM .But, K being prime, K 6= IM .Thus, if√

(N1 : IM ) 6√

(K : IM ) and√(N2 : IM )

√(K : IM ) gives N1 6 K. Thus, every primary element

containing N1 ∧ N2 is meet irreducible and hence, pradM (N1 ∧ N2) =pradM (N1) ∧ pradM (N2). �

Let, N be a proper element of an L-module M. Let p be a prime elementof L. For each positive integer n, we shall denote by K(N, pn),

K(N, pn) = ∨{A ∈M | aA 6 N ∨ pnIM for some a p}

Theorem 3.3. Let N be an element of an L-module M and p be a primeelement of L. For each positive integer n, K(N, pn) = IM or K(N, pn) is ap-primary element of M.

Proof. Assume that, K(N, pn) 6= IM .Let Q be an element of M properlycontaining K(N, pn).

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Let r 6 (K(N, pn) : Q). This implies, rQ 6 K(N, pn). Since, K(N, pn) 6Q ,there is some T 6 Q but T K(N, pn).Now,rT 6 K(N, pn).Therefore,thereis some c p such that c(rT ) 6 N ∨ PnIM . If r p then cr p andthis implies, T 6 K(N, pn) a contradiction. Hence, r 6 p.Which showsthat, [K(N, pn) : Q] 6 p. As p is prime

√[K(N, pn) : Q] 6 p.Now,we show

that,√

[K(N, pn) : IM ] = p.Let r 6√

[K(N, pn) : IM ].So,rtIM 6 K(N, pn)for some t ∈ Z+.Thus,c(rtIM ) 6 N ∨ pnIM for some c p.If r p

then crt p.This implies that IM 6 K(N, pn) a contradiction.Whichshows that,r 6 p and hence,

√[K(N, pn) : IM ] 6 p.Next,1pnIM =pnIM 6

N ∨ pnIM ,1 p.Hence,pnIM 6 K(N, pn) and so pn 6 [K(N.pn) : IM ] thatis p 6

√[K(N, pn) : IM ].This shows that,

√[K(N, pn) : IM ] = p.As,Q 6

IM ,√

[K(N, pn) : IM ] 6√

[K(N, pn) : Q] 6 p.Hence,√

[K(N, pn) : IM ] =√[K(N, pn) : Q] = p.Let rQ 6 K(N, pn) where,Q K(N, pn).As,Q

K(N, pn) as shown in first part r 6 p and hence,rkIM 6 K(N, pn) for someinteger k.Hence,K(N, pn) is p-primary element of M. �

The following theorem gives the description of the primary radical of anelement of a lattice module.

Theorem 3.4. Let N be an element of an L-module M. Then, pradM (N) =∧{K(N, pn) | p is prime element ofL, for some n ∈ Z+}.

Proof. K(N, pn) is primary element containing N. Therefore pradM (N) 6∧{K(N, pn) | p is prime element of L and for some n ∈ Z+}.For ev-ery primary element H in M such that N 6 H with p =

√(H : IM ) ∃

a positive integer r such that K(N, pr) 6 H. Therefore, ∧{K(N, pn) |p is prime element of L , for some n ∈ Z+}.6 K(N, pr) 6 H.Hence,∧{K(N, pr) | p is prime element of L, for some n ∈ Z+} 6 pradM (N).

�

Defination 1. A proper element N of a L-module M with pradM (N) = N

is called p-radical element.

The next result describes the relation between the p-radical elements ofM and the ascending chain condition for the same.

Theorem 3.5. Let M be an L-module. if M satisfies the ACC for p-radicalelements of M then every p-radical element of M is the meet of a finitenumber of primary elements.

Proof. Let N be p-radical element of M. Then pradM (N) = ∧i∈I{Ni |

Ni is primary element and N 6 Ni} = N . Assume that, I is the infi-nite index set and the expression N = ∧

i∈INi is reduced. Without loss of

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generality, we assume that, I is countable. Then, N =∞∧i=1

Ni 6∞∧i=2

Ni 6∞∧i=3

Ni 6 ................. which is an ascending chain of p-radical elements be-cause by theorem (3.1), ∧

iNi 6 pradM (∧

iNi) 6 ∧(pradM (Ni)) 6 ∧Ni. By

hypothesis, the ascending chain terminates. Therefore, ∃ a positive inte-ger j ∈ I such that

∞∧i=j

Ni =∞∧

i=j+1Ni =

∞∧

i=j+2Ni = .................. where,

∞∧

i=j+1Ni 6 Nj which contradicts the fact that the expression N = ∧

iNi is

reduced. Therefore, I must be finite.�

References

[1] Khouja Al,Eman A Maximal elements and prime elements in lattice modules. Dam-ascus university journal for basic sciences Vol 19, No 2, 203.

[2] Lamis J M Abulebda, The primary radical of a submodule, Advances in pure math-ematics, 2012, 2, 344 - 348

[3] Tavallaee Hamid A, Semi-radicals of sub modules in modules.IUST International Jour-nal of Engineering Science,Vol.19,No.1-2,2008,Page 21-27.

C S ManjarekarDepartment of MathematicsShivaji UniversityKolhapurcsmanjrekar@yahoo.co.in+91 9373 5615 30 / +91 9611 6879 25

U N KandaleGeneral Engineering DepartmentSharad Institute of Technology - College of EngineeringYadravujwalabiraje@gmail.com+91 8421 360061 / +91 9 6655 9 2207

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